4-manifold with $w_1\neq 0$, $w_1^2=0$, $w_2\neq 0$

Question

I wonder if there is a 4-manifold whose Stiefel-Whitney classes satisfy $w_1\neq 0$, $w_2\neq 0$, and $w_1^2=0$?

There is no 3-manifold whose Stiefel-Whitney classes are given by the above. For $\mathbb{R}P^4$, $w_1^2\neq 0$.

Answer 1

Let $K$ denote the Klein bottle.

As $K$ is non-orientable, $w_1(K) \neq 0$, but as $K$ is nullcobordant, $w_1(K)^2 = 0$ and $w_2(K) = 0$. Alternatively, as $K$ is a surface, $0 = \nu_2(K) = w_1(K)^2 + w_2(K)$ and $w_2(K)$ is the mod $2$ reduction of the Euler class; as $\chi(K) = 0$ is even, $w_2(K) = 0$ and therefore $w_1(K)^2 = 0$. Either way, we have $w_1(K\times S^2) = w_1(K) \neq 0$, $w_1(K\times S^2)^2 = w_1(K)^2 = 0$, and $w_2(K\times S^2) = 0$.

Finally, as $w_i(X\# Y) = w_i(X) + w_i(Y)$ for $i \neq 0$, we see that $M = (K \times S^2) \# \mathbb{CP}^2$ satisfies $w_1(M) = w_1(K) \neq 0$, $w_1(M)^2 = w_1(K)^2 = 0$ and $w_2(M) = w_2(\mathbb{CP}^2) \neq 0$.

Answer 2

Let $M$ be the non-orientable $S^3$ bundle over $S^1$. Its $\Bbb Z/2$ Betti numbers are $b_1 = 1, b_2 = 0, b_3 = 1$. It's non-orientable, so $w_1 \neq 0$, but clearly $w_1^2 = 0$. Now take $M \# \Bbb{CP}^2$. $\Bbb{CP}^2$ has $w_1 = 0$ but $w_2 \neq 0$.